I understand how they calculated NPW but how did they use the linear interpolation method?

I have never before seen their version of the linear interpolation method, but it does work. In linear interpolation we fit a linear function ##f = a + bx## through two points ##(x_1,f_1)## and ##(x_2,f_2)##:
[tex] f(x) = f_1 + \frac{f_2-f_1}{x_2-x_1} (x-x_1)[/tex]
This gives us the value of ##x_0##, the root of ##f(x) = 0,## as follows:
[tex] x_0 = \frac{f_2 x_1 - f_1 x_2}{f_2 - f_1}[/tex]
It follows (doing a lot of algebraic simplification) that
[tex] \frac{x_0 - x_1}{x_2-x_0} = -\frac{f_1}{f_2} [/tex]
In other words, we can find ##x_0## by solving the equation
[tex] \frac{x - x_1}{x_2-x} = -\frac{f_1}{f_2} [/tex]
The solution is ##x = x_0##. That is not how I would do it, but it is correct.

I understand how they calculated NPW but how did they use the linear interpolation method?

Just so you know: there is a difference between using linear interpolation on ##r## directly, and using linear interpolation on ##x = 1/(1+r)##. The present value is a polynomial in ##x##, but not in ##r##. Linear interpolation on ##x## gives ##x_0 \approx .8609359558##, giving ##r_0 \approx .1615265843##, or about 16.15%. Which answer is closer? Well, the exact solution is ##r = .1594684085##, or about 15.95%. So, in this case at least, interpolation on ##x## produces slightly better results.